codeforces B. Tape

B. Tapetime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou have a long stick, consisting of mm

segments enumerated from 11

to mm

. Each segment is 11

centimeter long. Sadly, some segments are broken and need to be repaired.You have an infinitely long repair tape. You want to cut some pieces from the tape and use them to cover all of the broken segments. To be precise, a piece of tape of integer length tt

placed at some position ss

will cover segments s,s+1,…,s+t−1s,s+1,…,s+t−1

.You are allowed to cover non-broken segments; it is also possible that some pieces of tape will overlap.Time is money, so you want to cut at most kk

continuous pieces of tape to cover all the broken segments. What is the minimum total length of these pieces?InputThe first line contains three integers nn

, mm

and kk

(1≤n≤1051≤n≤105

, n≤m≤109n≤m≤109

, 1≤k≤n1≤k≤n

) — the number of broken segments, the length of the stick and the maximum number of pieces you can use.The second line contains nn

integers b1,b2,…,bnb1,b2,…,bn

(1≤bi≤m1≤bi≤m

) — the positions of the broken segments. These integers are given in increasing order, that is, b1<b2<…<bnb1<b2<…<bn

.OutputPrint the minimum total length of the pieces.ExamplesInputCopy4 100 2
20 30 75 80
OutputCopy17
InputCopy5 100 3
1 2 4 60 87
OutputCopy6
NoteIn the first example, you can use a piece of length 1111

to cover the broken segments 2020

and 3030

, and another piece of length 66

to cover 7575

and 8080

, for a total length of 1717

.In the second example, you can use a piece of length 44

to cover broken segments 11

, 22

and 44

, and two pieces of length 11

to cover broken segments 6060

and 8787

.

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
long long int n[100010];
long long int m[100010];
int main()
{
    long long int a,b,c,t,i;
    cin >> a >> b >> c;
    for(i=1;i<=a;i++)
    {
        cin >> n[i];
    }
    for(i=1;i<=a-1;i++)
    {
        m[i]=n[i+1]-n[i]+1;
    }
    sort(m+1,m+a);
    t=n[a]-n[1]+1;
    for(i=a-1;i>a-c;i--)
    {
        t=t-m[i];
    }
    cout << t+((c-1)*2);
    return 0;
}